The heights of adult Virgin Islands men vary according to a normal distribution with:

mean m = 70 inches

and standard deviation s = 3 inches

Let Y be the random variable representing the heights of adult Virgin Islands men.

1 foot = 12 inches

=> 6 feet = 6 * 12 inches = 72 inches

5 feet 7 inches = (5 * 12 + 7) inches = 67 inches

6 feet 1 inches = (6 * 12 + 1) inches  = 73 inches

a) To find the proportion of men less than 72 inches tall, first we need to transform Y to the standard normal variable using the following transformation:

z = (Y - m)/s

=> P(Y < 72)

= P(z < (72 - 70)/3)

= P(z < 2/3)

= P(z < 0.67)

From the standard normal table, the area under the curve for z < 0.67 = 0.748 

Therefore, 0.748 or 74.8% of men in Virgin Islands are less than 6 feet tall.

b) To find the proportion of men between 67 inches and 73 inches tall, we transform Y to standard normal as:

P(67 < Y < 73)

= P((67 - 70)/3 < z < (73 - 70)/3)

= P(-3/3 < z  < 3/3)

= P(-1 < z < 1)

From the standard normal table, the area under the curve for -1 < z < 1

= Area under the curve for z < 1  - area under the curve for z < -1

= 0.841 - 0.159

= 0.682 [Ans]

Therefore, 0.682 or 68.2% of men in Virgin Islands are between 5 feet 7 inches and 6 feet 1 inches tall.